  Question

50 mL of a solution of Na2CO3 and NaHCO3 requires 17.5 mL of 0.1 M H2SO4 for neutralisation using phenolphthalein as indicator. Methyl orange is then added when further 25 mL of 0.2 M H2SO4 was required. Calculate the weight of Na2CO3 and NaHCO3 respectively in the solution.

A
1.454 g, 0.221 g  B
0.667 g, 0.118 g  C
0.892 g, 1.651 g  D
0.371 g, 0.546 g  Solution

The correct option is D 0.371 g, 0.546 gPhenolphthalein will only neutralise 50% of Na2CO3. 12 Milliequivalents of Na2CO3=Milliequivalents of H2SO4=0.1×17.5×2=3.5 ∴ Milliequivalents of Na2CO3=7.0 WNa2CO3MNa2CO3×2×1000=7.0 ∴ WNa2CO3=7.0×1062×1000=0.371 g The methyl orange indicator will neutralise the rest 50% of Na2CO3 and original NaHCO3. ∴ 12 Milliequivalents of Na2CO3+Milliequivalents of NaHCO3=Milliequivalents of H2SO4=0.2×25×2=10 ∴ Milliequivalents of NaHCO3=10−3.5=6.5 WNaHCO3MNaHCO3×1×1000=6.5 ∴ WNaHCO3=6.5×841×1000=0.546 g  Chemistry

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